Number is divisible by 29 or not
Given a large number n, find if the number is divisible by 29.
Examples :
Input : 363927598 Output : No Input : 292929002929 Output : Yes
A quick solution to check if a number is divisible by 29 or not is to add 3 times of last digit to rest number and repeat this process until number comes 2 digit. The given number is divisible by 29 iff the obtained two digit number is divisible by 29.
Number is 348,
Three times of last digit + Rest of the number = 8*3 + 34 = 58
Since 58 is divisible by 29, 348 is also divisible by 29.
C++
// CPP program to demonstrate above method // to check divisibility by 29. #include <iostream> using namespace std; // Returns true if n is divisible by 29 // else returns false. bool isDivisible(long long int n) { // add the lastdigit*3 to renaming // number until number comes only // 2 digit while (n / 100) { int last_digit = n % 10; n /= 10; n += last_digit * 3; } // return true if number is // divisible by 29 another return (n % 29 == 0); } // Driver Code int main() { long long int n = 348; if (isDivisible(n)) cout << "Yes" << endl; else cout << "No" << endl; return 0; } |
chevron_right
filter_none
Java
// Java program to demonstrate above method // to check divisibility by 29. import java.io.*; class GFG { // Returns true if n is divisible by 29 // else returns false. static boolean isDivisible(long n) { // add the lastdigit*3 to renaming // number until number comes only // 2 digit while (n / 100 > 0) { int last_digit = (int)n % 10; n /= 10; n += last_digit * 3; } // return true if number is // divisible by 29 another return (n % 29 == 0); } // Driver code public static void main(String[] args) { long n = 348; if (isDivisible(n)) System.out.println("Yes"); else System.out.println("No"); } } // This code is contributed by vt_m. |
chevron_right
filter_none
Python3
# Python3 program to demonstrate above # method to check divisibility by 29. # Returns true if n is divisible # by 29 else returns false. def isDivisible(n): # add the lastdigit*3 to renaming # number until number comes only # 2 digit while (int(n / 100)) : last_digit = int(n % 10) n = int(n / 10) n += last_digit * 3 # return true if number is # divisible by 29 another return (n % 29 == 0) # Driver Code n = 348 if(isDivisible(n) != 0): print("Yes") else: print("No") # This code is contributed by Smitha Dinesh Semwal. |
chevron_right
filter_none
C#
// C# program to demonstrate above method // to check divisibility by 29. using System; class GFG { // Returns true if n is divisible by 29 // else returns false. static bool isDivisible(long n) { // add the lastdigit*3 to renaming // number until number comes only // 2 digit while (n / 100 > 0) { int last_digit = (int)n % 10; n /= 10; n += last_digit * 3; } // return true if number is // divisible by 29 another return (n % 29 == 0); } // Driver code public static void Main() { long n = 348; if (isDivisible(n)) Console.Write("Yes"); else Console.Write("No"); } } // This code is contributed by nitin mittal |
chevron_right
filter_none
PHP
<?php // PHP program to demonstrate // above method to check // divisibility by 29. // Returns true if n is // divisible by 29 // else returns false. function isDivisible($n) { // add the lastdigit*3 to // remaining number until // number becomes of only // 2 digit while (intval($n / 100)) { $last_digit = $n % 10; $n = intval($n / 10); $n += $last_digit * 3; } // return true if number is // divisible by 29 another return ($n % 29 == 0); } // Driver Code $n = 348; if (isDivisible($n)) echo "Yes"; else echo "No" ; // This code is contributed by Sam007 ?> |
chevron_right
filter_none
Output :
Yes
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.
